So we know that you use a 26-letter wheel with letters arranged in a random (by a human or a computer - what method was used to pick the numbers?) order, and that the order in which the letters are read could be significant. Solvable by a computer? Eventually, yes. Solvable by a human? Most probably not.

Re: I have devised an uncrackable substitution cypher code

Re: I have devised an uncrackable substitution cypher code

Kerckhoffs' principle

In cryptography, Kerckhoffs' principle (also called Kerckhoffs' assumption, axiom or law) was stated by Auguste Kerckhoffs in the 19th century: a cryptosystem should be secure even if everything about the system, except the key, is public knowledge.

Kerckhoffs' principle was reformulated (perhaps independently) by Claude Shannon as "The enemy knows the system." In that form, it is called Shannon's maxim. In contrast to "security through obscurity," it is widely embraced by cryptographers.

Show the algorithm but not the key. if it can be reconstructed without the key, it is breakable. i doubt that i can do it, but a really good cryptanalyst can probably do it

Visit calccrypto.wikidot.com for detailed descriptions of algorithms and other crypto related stuff (not much yet, so help would be appreciated).

Re: I have devised an uncrackable substitution cypher code

I am not familiar with the terms, especially key. Hidden within the code is instructions on where to start, which direction to go, and details of the gaps between encrypted letters. Only the recipient knows which of the 900 letters in the code give these instructions. For example, a particular letter in the coded text would tell you which ROW to start in, so A=1, B=2 etc etc. Where to start in the row is more cunningly coded. The gaps between encoded letters are also defined in particular cells of the matrix.

I am not being difficult, folks, but you have to take any letter, ie 900 choices, then any other, ie 899 and so on.

So for a 20 word message that is 900 x 899 x 898..................x 880 =X

Then for EACH of these strings, you have to try all 26 letters of the alphabet.

So the total number of tries is X multiplied by 20 to the power 26

Within your tries you will find many many many words that string together.

I could have encoded the words in a different order to make it more difficult.......

I do not know how to calculate the total number of permutations, but 20 to the power 26 is pretty big......in itself.

Re: I have devised an uncrackable substitution cypher code

This all depends on the amount of "waffle". The way I would do it normally would be to take the first, 27th, 53rd etc and do a frequential analysis of those to determine the substitution for those, then repeat that for the 2nd, 28th, 54th and so on. However, the "waffle" would throw this analysis off; a little would probably still work because it would only affect the proportions marginally. However lots would seriously mess up the proportions (though it would still be breakable- just harder to get frequency analysis right. The most common would still correspond to "e", but then there's the trouble of working out what's waffle).

If you want a tough code, try this one:ÍÐÇÆ|vr¸ºin||vd²»¿it¥ÎÄ²gboµr{xÍtwiÁÂÉÂÍÆ»if{{}eµº¾¼qtmÈÄÄ®µµÄ§`Woo¡ªß¦xd´opt»º³lt}«`\|Ã´»»xqqÇy£rrpyÊÖ¨¨}}reµº¾¼»»ÈÄÄ®µµÄ¹¼¹ÍÒÄ¹¸¥}¢Á|noµrmI¢¢ÎÃxyn»³pn}}vg½ÏÏÍ